Let α, β and γ be real numbers. Consider the following system of linear equationsx + 2y + z = 7x + αz = 112x – 3y + βz = γMatch each entry in List-I to the correct entries in List-II. List-I List-II P)If β=12(7α−3) and γ=28 then the system has1)a unique solutionQ)If β=12(7α−3) and γ≠28, then the system has2)no solutionR)If β≠12(7α−3), where α = 1 and γ≠28, then the system has3)infinitely many solutionsS)If β≠12(7α−3) where α = 1 and γ=28, then the system has4)x = 11, y = –2 and z = 0 as a solution 5)x = –15, y = 4 and z = 0 as a solution PQRS1)32142)32543)21454)2113

x+2y+z=7 ----(1) x+αz=11 ----(2) 2x−3y+βz=γ ---(3)3×(1)+2×(3)⇒7x+z(3+2β)=21+2γ If 71=3+2βα=21+2γ11⇒7α=3+2β⇒β=12(7α−3) and 77=21+2γ⇒r=28If β=12(7α−3) and γ=28 then the system has infinitely many solutions ⇒P→3If β=12(7α−3) and γ≠28, then the system has no solution ⇒Q→2If β≠12(7α−3), where α = 1 and γ≠28, then the system has unique solution ⇒R→1If β≠12(7α−3) where α = 1 and γ=28, then x = 11, y = -2, z = 0 is a solution to the given system S→4.

Let α, β and γ be real numbers. Consider the following system of linear equationsx + 2y + z = 7x + αz = 112x – 3y + βz = γMatch each entry in List-I to the correct entries in List-II. List-I List-II P)If β=12(7α−3) and γ=28 then the system has1)a unique solutionQ)If β=12(7α−3) and γ≠28, then the system has2)no solutionR)If β≠12(7α−3), where α = 1 and γ≠28, then the system has3)infinitely many solutionsS)If β≠12(7α−3) where α = 1 and γ=28, then the system has4)x = 11, y = –2 and z = 0 as a solution 5)x = –15, y = 4 and z = 0 as a solution PQRS1)32142)32543)21454)2113

x+2y+z=7 ----(1) x+αz=11 ----(2) 2x−3y+βz=γ ---(3)3×(1)+2×(3)⇒7x+z(3+2β)=21+2γ If 71=3+2βα=21+2γ11⇒7α=3+2β⇒β=12(7α−3) and 77=21+2γ⇒r=28If β=12(7α−3) and γ=28 then the system has infinitely many solutions ⇒P→3If β=12(7α−3) and γ≠28, then the system has no solution ⇒Q→2If β≠12(7α−3), where α = 1 and γ≠28, then the system has unique solution ⇒R→1If β≠12(7α−3) where α = 1 and γ=28, then x = 11, y = -2, z = 0 is a solution to the given system S→4.

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Let α, β and γ be real numbers. Consider the following system of linear equations
x + 2y + z = 7
x + αz = 11
2x – 3y + βz = γ
Match each entry in List-I to the correct entries in List-II.
List-I List-II
P) If $β = \frac{1}{2} (7 α - 3)$ and $γ = 28$ then the system has 1) a unique solution
Q) If $β = \frac{1}{2} (7 α - 3)$ and $γ \neq 28$ , then the system has 2) no solution
R) If $β \neq \frac{1}{2} (7 α - 3)$ , where $α = 1$ and $γ \neq 28$ , then the system has 3) infinitely many solutions
S) If $β \neq \frac{1}{2} (7 α - 3)$ where $α = 1$ and $γ = 28$ , then the system has 4) x = 11, y = –2 and z = 0 as a solution
5) x = –15, y = 4 and z = 0 as a solution
P Q R S
1) 3 2 1 4
2) 3 2 5 4
3) 2 1 4 5
4) 2 1 1 3

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Detailed Solution

$x + 2 y + z = 7 - - - - (1) x + αz = 11 - - - - (2) 2 x - 3 y + βz = γ - - - (3)$
$\begin{array}{l} 3 \times (1) + 2 \times (3) \Rightarrow 7 x + z (3 + 2 β) = 21 + 2 γ If \frac{7}{1} = \frac{3 + 2 β}{α} = \frac{21 + 2 γ}{11} \\ \Rightarrow 7 α = 3 + 2 β \Rightarrow β = \frac{1}{2} (7 α - 3) and 77 = 21 + 2 γ \Rightarrow r = 28 \end{array}$
If $β = \frac{1}{2} (7 α - 3)$ and $γ = 28$ then the system has infinitely many solutions $\Rightarrow P \to 3$
If $β = \frac{1}{2} (7 α - 3)$ and $γ \neq 28$ , then the system has no solution $\Rightarrow Q \to 2$
If $β \neq \frac{1}{2} (7 α - 3)$ , where $α = 1$ and $γ \neq 28$ , then the system has unique solution $\Rightarrow R \to 1$
If $β \neq \frac{1}{2} (7 α - 3)$ where $α = 1$ and $γ = 28$ , then x = 11, y = -2, z = 0 is a solution to the given system $S \to 4$ .

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